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# 1.2: PDE’s

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Rather than giving a strict mathematical definition, let us look at an example of a PDE, the heat equation in 1 space dimension

$\dfrac{\partial^2 u(x,t)}{\partial x^2} = \frac{1}{k} \dfrac{\partial u(x,t)}{\partial t}. \label{eq:I:heat}$

• It is a PDE since partial derivatives are involved.

Review

To remind you of what that means: $$\dfrac{\partial}{\partial x}{u(x,t)}$$ denotes the differentiation of $$u(x,t)$$ w.r.t. $$x$$ keeping $$t$$ fixed,

$\dfrac{\partial}{\partial x} (x^2t+xt^2) = 2xt+t^2.$

• Equation \ref{eq:I:heat} called linear since $$u$$ and its derivatives appear linearly, i.e., once per term. No functions of $$u$$ are allowed. Terms like $$u^2$$, $$\sin(u)$$, $$u\dfrac{\partial}{\partial x}{u}$$, etc., break this rule, and lead to non-linear equations. These are interesting and important in their own right, but outside the scope of this course.
• Equation \ref{eq:I:heat} is also homogeneous (which just means that every term involves either $$u$$ or one of its derivatives, there is no term that does not contain $$u$$). The equation $\dfrac{\partial^2}{\partial x^2}{u(x,t)} = \frac{1}{k} \dfrac{\partial}{\partial t}{u(x,t)}+\sin(x)$ is called inhomogeneous, due to the $$\sin(x)$$ term on the right, that is independent of $$u$$.

Linearity

Why is all that so important? A linear homogeneous equation allows superposition of solutions. If $$u_1$$ and $$u_2$$ are both solutions to the heat equation,

$\dfrac{\partial^2}{\partial x^2}{u_1(x,t)} - \frac{1}{k} \dfrac{\partial}{\partial t}{u_1(x,t)}= \dfrac{\partial}{\partial t}{u_2(x,t)} - \frac{1}{k} \dfrac{\partial^2}{\partial x^2}{u_2(x,t)}=0, \label{eq:I:heatsol}$

any combination is also a solution,

$\dfrac{\partial^2}{\partial x^2}{[a u_1(x,t)+bu_2(x,t)]} - \frac{1}{k} \dfrac{\partial}{\partial t}{[au_1(x,t)+b u_2(x,t)]}=0.$

For a linear inhomogeneous equation this gets somewhat modified. Let $$v$$ be any solution to the heat equation with a $$\sin(x)$$ inhomogeneity,

$\dfrac{\partial^2}{\partial x^2}{v(x,t)} - \frac{1}{k} \dfrac{\partial}{\partial t}{v(x,t)}=\sin(x).$

In that case $$v+au_1$$, with $$u_1$$ a solution to the homogeneous equation, see Equation \ref{eq:I:heatsol}, is also a solution,

\begin{aligned} \dfrac{\partial^2}{\partial x^2}{[v(x,t)+a u_1(x,t)]} - \frac{1}{k} \dfrac{\partial}{\partial t}{[v(x,t)+a u_1(x,t)]}&=&\nonumber\\ \dfrac{\partial^2}{\partial x^2}{v(x,t)} - \frac{1}{k} \dfrac{\partial}{\partial x}{v(x,t)} +a \left(\dfrac{\partial}{\partial x}{u_1(x,t)} - \frac{1}{k} \dfrac{\partial}{\partial t}{u_1(x,t)}\right)&=& \sin(x).\end{aligned}

Finally we would like to define the order of a PDE as the power in the highest derivative, even it is a mixed derivative (w.r.t. more than one variable).

Exercise $$\PageIndex{1}$$

Which of these equations is linear? and which is homogeneous?

1. $$\dfrac{\partial^2 u}{\partial x^2} + x^2 \dfrac{\partial u}{\partial y} = x^2 + y^2$$
2. $$y^2\dfrac{\partial^2 u}{\partial x^2}+u\dfrac{\partial u}{\partial x} + x^2\dfrac{\partial^2 u}{\partial y^2} = 0$$
3. $$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial x^2}=0$$
Exercise $$\PageIndex{2}$$
1. $$\dfrac {\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0$$
2. $$\dfrac{\partial^2 u}{\partial x^2} - 2\dfrac{\partial^4 u}{\partial^3 x \partial y} + \dfrac{\partial^2 u}{\partial y^2}= 0$$